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(A) The resistance of a wire is given by $R = \rho \frac{L}{A}$. Since the material $(\rho)$ and length $(L)$ are the same, $R \propto \frac{1}{A}$. T

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In series, the fine wire liberates more energy, while in parallel, the thick wire liberates more energy.

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(A) The resistance of a wire is given by $R = \rho \frac{L}{A}$. Since the material $(\rho)$ and length $(L)$ are the same, $R \propto \frac{1}{A}$. Thus, the fine wire has a smaller cross-sectional area $(A)$ and higher resistance $(R)$, while the thick wire has a larger area $(A)$ and lower resistance $(R)$.$1$. In series connection, the current $(I)$ is the same through both coils. The power dissipated is $P = I^2 R$. Since $P \propto R$, the coil with higher resistance (the fine wire) will liberate more energy.$2$. In parallel connection, the potential difference $(V)$ is the same across both coils. The power dissipated is $P = \frac{V^2}{R}$. Since $P \propto \frac{1}{R}$, the coil with lower resistance (the thick wire) will liberate more energy.

Two heating coils, one of fine wire and the other of thick wire of the same material and of the same length, are connected in series and in parallel. Which of the following statements is correct?

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